Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

二叉查找树的特性（左节点值小于根小于右节点值）决定了同父节点的值n和任意两个节点p，q之间一定满足关系 p.val < root.val < q.val

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    TreeNode cur = root;
    if (cur.val > p.val && cur.val < q.val){
        return cur;
    } else if(cur.val > p.val && cur.val > q.val){
        return lowestCommonAncestor(root.left, p, q);
    } else if(cur.val < p.val && cur.val < q.val){
        return lowestCommonAncestor(root.right, p, q);
    }
    return root;
}

Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

如果root为根节点的话，可以用divide & conque 遍历解决。时间复杂度 O(n)

public static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    if (root == null || root.val == p.val || root.val == q.val) {
        return root;
    }

    // Divide
    TreeNode left = lowestCommonAncestor(root.left, p, q);
    TreeNode right = lowestCommonAncestor(root.right, p, q);

    // Conquer
    if (left != null && right != null) {
        return root;
    }
    if (left != null) {
        return left;
    }
    if (right != null) {
        return right;
    }
    return null;
}